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2 years ago

PLEASE HELP ME ASAP??!!

Mathematics

1 answer:

earnstyle [38]2 years ago

5 0

It says that there was an error with my answer, but I don't know what, so I screenshotted what I typed and attached them as 4 images below:

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Which of the following is an example of the Distributive Property?

In-s [12.5K]

Answer:

2z2(2y - 5) = 4z2y - 10z2

Step-by-step explanation:

I hope this helps

8 0

3 years ago

Read 2 more answers

3 Quiz

sertanlavr [38]

Answer:

y varies of directly as x. This means that the equation is of the form y=kx where k is any number.

Substituting y=12 when x=4, we get

12=k×4⇒k=3

Hence, the required equation is y=3x.

The value of y when x=5 is y=3x5=15.

Step-by-step explanation:

bro it will work ♥️

5 0

3 years ago

Let g represent Mia’s score. Which expression represents 57 more than 3 times Mia’s score?

My name is Ann [436]

Answer

Step-by-step explanation:

3g+57

6 0

3 years ago

Find 3x2 − y3 − y3 − z if x = 3, y = −2, and z = −5.

IceJOKER [234]

3x²-y³ - y³ - z if x = 3, y = -2, z = -5

Simply plug in all the values :)

3(3²) - (-2³) - (-2³) - (-5)

Simplify.

3(9) + 2³ + 2³ + 5

Simplify.

27 + 8 + 8 + 5

Simplify.

35 + 13

Simplify.

48

~Hope I helped!~

4 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

2 years ago